Current cut-off Circuit

Thread Starter

Jess Beca

Joined Oct 28, 2017
15
Hey, i would like some assistance with my circuit. I need to design a current cut-off circuit that will disconnect when the current coming into the circuit exceeds 100mA. Also, i would are there any ideas of designing a voltage limiter circuit, it should not allow the voltage to be more than 3.3volts, these are protection circuits for a cold.
 

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xox

Joined Sep 8, 2017
838
Ever heard of a "crowbar"? Limiting the voltage should prevent over-current in most types of circuits too, though I'm not sure about the particular one you're working on (sorry, I'm just a newb, no circuit analysis skills whatsoever).
 

#12

Joined Nov 30, 2010
18,224
A crowbar circuit is a sort of suicidal method because it takes out a fuse and stops everything. The LM723 chip is equipped with a foldback circuit which greatly diminishes the output voltage during an overload and recovers as soon as the short is removed. I fell in love with that circuit back in the 1970's because it's almost* idiot proof and causes no damage.

*Yes, I know: If you try to make it idiot proof, all you accomplish is a better idiot detector.:(
 

Thread Starter

Jess Beca

Joined Oct 28, 2017
15
Hey, thanks xox and #12 for the replies.

#12
Is there a similar, newer device that will do a better job. I really don't wan tot use an idiot proof, since its an electronic assignment. Also, i forgot to state that these protection circuit will be connected to a AC to DC converter.
 

AlbertHall

Joined Jun 4, 2014
12,345
A crowbar circuit is a sort of suicidal method because it takes out a fuse and stops everything.
But if you already have a circuit which limits the current to a safe value which the TS circuit is trying to do then a crowbar is a good solution for overvoltage. But it would be wise to include a fuse as well just in case.
 

#12

Joined Nov 30, 2010
18,224
D'oh. I forgot for a moment that a crowbar is for over VOLTAGE and a foldback is for over CURRENT.:oops:
 

Thread Starter

Jess Beca

Joined Oct 28, 2017
15
Hi, AlbertHall
Thanks for the reply, i understand what you saying, but how would modify my circuit to include the crowbar with the fuse? Also, can i use a relay instead of a fuse?
 

Reloadron

Joined Jan 15, 2015
7,501
Hey Ron,
Thanks for the reply, can how would you design the comparator circuit? Taken from the other thread.

I would use the ACS 712 unit I linked to for current. Now for the voltage I would likely use a basic compactor circuit. A simple Google will yield dozens of comparators. Anyway you get a condition for over current or Over Voltagethose signals could feed a NAND gate so if either one exceeds a limit a relay shuts down your power.

Ron
 

AlbertHall

Joined Jun 4, 2014
12,345
Hi, AlbertHall
Thanks for the reply, i understand what you saying, but how would modify my circuit to include the crowbar with the fuse? Also, can i use a relay instead of a fuse?
Add the crowbar between the current limited supply and your load. You will find plenty of circuits on www.
A fuse is quick a relay isn't. I have never known a fuse to fail to blow but a relay circuit may fail and you wouldn't know about it until something smokes.
 

Picbuster

Joined Dec 2, 2013
1,047
Hey, i would like some assistance with my circuit. I need to design a current cut-off circuit that will disconnect when the current coming into the circuit exceeds 100mA. Also, i would are there any ideas of designing a voltage limiter circuit, it should not allow the voltage to be more than 3.3volts, these are protection circuits for a cold.
use a (bourns) 100mA multi fuse.
The impedance goes extreme high when the 100mA is reached.
Voltage use zener and P mosfet

Picbuster
 

Thread Starter

Jess Beca

Joined Oct 28, 2017
15
Hey, Ron, I decided to use a window comparator for the voltage limit, could you please check the attached circuit. The limit would be 0 - 3.3 volts, and i want achieve a 3.1 volts output.
 

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Reloadron

Joined Jan 15, 2015
7,501
Hey, Ron, I decided to use a window comparator for the voltage limit, could you please check the attached circuit. The limit would be 0 - 3.3 volts, and i want achieve a 3.1 volts output.
When doing a comparator circuit your best bet is to use a chip designed as a comparator, Chips like the LM339 seines are a much better choice especially considering a single ended supply. Chips like this will work much better for your application.

Ron
 

Thread Starter

Jess Beca

Joined Oct 28, 2017
15
I’m trying to achieve a 0 to 3.1 volts peak to peak, but I want to set the limit to 3.3 and the reference voltage as 3.1. If the voltage exceeds the 3.3 the LED turns on.
 

Reloadron

Joined Jan 15, 2015
7,501
Okay, but is the circuit design correctly? Any suggestion for the resistor values?
At a glance your design looks fine. Keep something in mind. The chip I suggested is an "open collector" output.
An open collector is a common type of output found on many integrated circuits (IC), which behaves like a switch that is either connected to ground or disconnected.

Instead of outputting a signal of a specific voltage or current, the output signal is applied to the base of an internal NPN transistor whose collector is externalized (open) on a pin of the IC. The emitter of the transistor is connected internally to the ground pin.

Also keep in mind that the output current (sink current) is not very high so if you wish to drive for example a small relay to drop out your supply it is wise to include a transistor at the comparator output. Your resistor(s) for your reference voltage are simply a voltage divider network. What voltages do you have available? The reason I ask is because you want a stable Vcc source. For example if you have 12 VDC available I would use a simple 5 volt regulator like a 7805 Regulator to power this circuit. The Vcc will determine you divider values.

I just saw your latest post where you mention Peak to Peak? I am assuming 3.3 VDC and peak to peak implies an AC voltage. Not sure what you have in mind with that?

Ron
 

Thread Starter

Jess Beca

Joined Oct 28, 2017
15
At a glance your design looks fine. Keep something in mind. The chip I suggested is an "open collector" output.
An open collector is a common type of output found on many integrated circuits (IC), which behaves like a switch that is either connected to ground or disconnected.

Instead of outputting a signal of a specific voltage or current, the output signal is applied to the base of an internal NPN transistor whose collector is externalized (open) on a pin of the IC. The emitter of the transistor is connected internally to the ground pin.

Also keep in mind that the output current (sink current) is not very high so if you wish to drive for example a small relay to drop out your supply it is wise to include a transistor at the comparator output. Your resistor(s) for your reference voltage are simply a voltage divider network. What voltages do you have available? The reason I ask is because you want a stable Vcc source. For example if you have 12 VDC available I would use a simple 5 volt regulator like a 7805 Regulator to power this circuit. The Vcc will determine you divider values.

I just saw your latest post where you mention Peak to Peak? I am assuming 3.3 VDC and peak to peak implies an AC voltage. Not sure what you have in mind with that?

Ron
Yes, it is an AC voltage, the current and voltage circuit are protection circuits for a cpld. The cpld will measure the frequency of the input signal, but the cpld cannot voltages greater than 3.1 and 1A. The cpld must measure input signals for voltage level 0 - 3.1V pk to pk (cannot exceeds 3.3). Also these circuit will be connected to a AC to DC converter that will go to the board. If the current exceeds the circuit must disconnect.
 
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Thread Starter

Jess Beca

Joined Oct 28, 2017
15
The voltage supply has be decided by me, I was thinking a 12 VDC, and as you suggested i would the 7805 Regulator to power the circuit.
 
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